In a cricket match, a player throws the ball to the wicketkeeper which is at 20 m horizontal distance. The player throws the ball with speed u at angle 45° with the horizontal. The value of u such that wicketkeeper just catches the ball directly, will be

Solution:

The horizontal distance covered by a projectile is called its range (R). The formula for the range of a projectile thrown with initial speed 'u' at an angle 'θ' with the horizontal is given by:

$R = \frac{u^2 \sin(2\theta)}{g}$

Given:

Horizontal distance, R = 20 m

Angle of projection, θ = 45°

Acceleration due to gravity, g ≈ 10 m/s² (standard approximation for such problems)

Substitute the given values into the formula:

$20 = \frac{u^2 \sin(2 \times 45°)}{10}$

$20 = \frac{u^2 \sin(90°)}{10}$

Since $\sin(90°) = 1$:

$20 = \frac{u^2 \times 1}{10}$

$200 = u^2$

$u = \sqrt{200}$

$u = \sqrt{100 \times 2}$

$u = 10\sqrt{2}$ m/s

Thus, the value of u is $10\sqrt{2}$ m/s.